= Solution
This is an infinite <binary expansion> with digits in $\{0,1\}$. If the usual expansion of $\alpha$ has infinitely many $1$ digits, list their positions as $b_1<b_2<\cdots$. If it terminates with a final term $2^{-k}$, replace that term by
$$
2^{-k}=\sum_{j=k+1}^{\infty}2^{-j};
$$
this also covers every <dyadic rational>. For $\alpha=1$, use $b_n=n$. Thus every $\alpha\in(0,1]$ has the required form.
The representation by an infinite strictly increasing sequence is unique. If two sequences first differ at exponent $k$, one sum contains $2^{-k}$ and the other does not. The latter's entire possible tail is at most $\sum_{j>k}2^{-j}=2^{-k}$, while the former has $2^{-k}$ plus its own nonempty tail, so the sums cannot agree.
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